Chapter 9, Problem 3SP

(a)

To determine

The volume of the block in cubic meters.

Answer to Problem 3SP

The volume of the block in cubic meters is $2.7\times {10}^{-5}{\text{ m}}^3$.

Explanation of Solution

Given info: The density of the steel block is $7800\text{kg}/{\text{m}}^3$. The side of the cubic block is $0.03\text{m}$.

Write the expression to find the volume of the block.

$V=a^3$

Here,

$V$ is the volume of the block

$a$ is the side of the cube

Substitute $0.03\text{m}$ for $a$ in the above equation.

$\begin{array}{c}V={\left(0.03\text{m}\right)}^3\\ =2.7\times {10}^{-5}{\text{ m}}^3\end{array}$

Conclusion:

Therefore, the volume of the block in cubic meters is $2.7\times {10}^{-5}{\text{ m}}^3$.

(b)

To determine

The mass of the block.

Answer to Problem 3SP

The mass of the block is $0.21\text{ kg}$.

Explanation of Solution

Write the expression to find the mass of the block when the density of the block is known.

$M=\rho V$

Here,

$\rho$ is the density of the block

$M$ is the mass of the block

Substitute $7800\text{kg}/{\text{m}}^3$ for $\rho$ and $2.7\times {10}^{-5}{\text{ m}}^3$ for $V$ in the above equation.

$\begin{array}{c}M=\left(7800\text{kg}/{\text{m}}^3\right)\left(2.7\times {10}^{-5}{\text{ m}}^3\right)\\ =0.21\text{ kg}\end{array}$

Conclusion:

Therefore, the mass of the block is $0.21\text{ kg}$.

(c)

To determine

The weight of the block.

Answer to Problem 3SP

The weight of the block is $2.06\text{N}$.

Explanation of Solution

Write the equation to find the weight of the block.

$W=Mg$

Here,

$W$ is the weight of the block

$M$ is the mass of the block

$g$ is the acceleration due to gravity

Substitute $0.21\text{ kg}$ for $M$ and $9.8\text{m}/{\text{s}}^2$ for $g$ to find the weight of the block.

$\begin{array}{c}W=\left(0.21\text{ kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\\ =2.06\text{N}\end{array}$

Conclusion:

Therefore, the weight of the block is $2.06\text{N}$.

(d)

To determine

The buoyant force acting on the block.

Answer to Problem 3SP

The buoyant force acting on the block is $0.265\text{}\text{ N}$.

Explanation of Solution

The buoyant force acting on the block is defined as equal to the weight of the water displaced.

Write the equation to find the buoyant force.

$\begin{array}{c}{F}_b=m_{w}g\\ ={\rho }_w{V}_{w}g\end{array}$

Here,

$F_b$ is the buoyant force

${\rho }_w$ is the density of water

$V_w$ is the volume of water displaced

$g$ is the acceleration due to gravity

Substitute $1000\text{kg}/{\text{m}}^3$ for ${\rho }_w$, $2.7\times {10}^{-5}{\text{ m}}^3$ for $V_w$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in the above equation to find the buoyant force.

$\begin{array}{c}{F}_{bouyant}=\left(1000\text{kg}/{\text{m}}^3\right)\left(2.7\times {10}^{-5}{\text{ m}}^3\right)\left(9.8\text{m}/{\text{s}}^2\right)\\ =0.265\text{}\text{ N}\end{array}$

Conclusion:

Therefore, the buoyant force acting on the block is $0.265\text{}\text{ N}$.

(e)

To determine

The tension in the string needed to hold the block in place.

Answer to Problem 3SP

The tension in the string needed to hold the block in place is $1.80\text{ N}$.

Explanation of Solution

Write the expression of Newton’s first law of motion using the total forces acting on the block to find the tension on the string.

$T=W-F_b$

Here,

$T$ is the tension on the string

$W$ is the weight of the block

$F_b$ is the buoyant force

Substitute $2.06\text{N}$ for $W$ and $0.265\text{}\text{ N}$ for $F_b$ in the above equation.

$\begin{array}{c}T=2.06\text{N}-0.265\text{}\text{ N}\\ =1.80\text{ N}\end{array}$

Conclusion:

Therefore, the tension in the string needed to hold the block in place is $1.80\text{ N}$.